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How would you use the substitution: $x=1+sin\theta$ to evaluate $\int_{0}^{π/2}\frac{cos\theta}{(1+sin\theta)^3}d\theta$.

Furthermore what would you when changing the limits, since its a definite integral.

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    $\begingroup$ Any tries or thoughts yourself? It is a somewhat simple example... $\endgroup$ – mickep Oct 25 '15 at 12:11
  • $\begingroup$ What would you do to find the answer. $\endgroup$ – DanJa78912 Oct 25 '15 at 12:16
  • $\begingroup$ Like what would you do to find the new limits. $\endgroup$ – DanJa78912 Oct 25 '15 at 12:20
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HINT:

$$\int_{0}^{\frac{\pi}{2}}\left(\frac{\cos(x)}{(1+\sin(x))^3}\right)\text{d}x=$$


Substitute $u=\sin(x)+1$ and $\text{d}u=\cos(x)\text{d}x$ and $ \begin{cases} x=0 \Longrightarrow u=\sin(0)+1=1\\\ x=\frac{\pi}{2} \Longrightarrow u=\sin\left(\frac{\pi}{2}\right)+1=2 \\ \end{cases} $:


$$\int_{1}^{2}\left(\frac{1}{u^3}\right)\text{d}u=$$ $$\int_{1}^{2}\left(u^{-3}\right)\text{d}u=$$ $$\left[-\frac{1}{2u^2}\right]_{1}^{2}=$$ $$-\frac{1}{2}\left[\frac{1}{u^2}\right]_{1}^{2}=$$ $$-\frac{1}{2}\left(\frac{1}{2^2}-\frac{1}{1^2}\right)=$$ $$-\frac{1}{2}\left(\frac{1}{4}-\frac{1}{1}\right)=$$ $$-\frac{1}{2}\left(\frac{1}{4}-1\right)=$$ $$-\frac{1}{2}\left(-\frac{3}{4}\right)=$$ $$--\frac{1}{2}\cdot\frac{3}{4}=$$ $$\frac{1}{2}\cdot\frac{3}{4}=$$ $$\frac{1\cdot 3}{2\cdot 4}=$$ $$\frac{3}{8}$$

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  • $\begingroup$ What would you do to find the new limits $\endgroup$ – DanJa78912 Oct 25 '15 at 12:19
  • $\begingroup$ @Elliot-S00 Look to my edit! $\endgroup$ – Jan Oct 25 '15 at 12:25
  • $\begingroup$ You forgot to make u=1+sinx $\endgroup$ – DanJa78912 Oct 25 '15 at 12:45
  • $\begingroup$ I didn't, look good to my answer! $\endgroup$ – Jan Oct 25 '15 at 12:48

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